Problem 38 - Thomas Calculus

Question

In Exercises 33–38, find the distance from the point to the line. $$ (-1,4,3) ; \quad x=10+4 t, \quad y=-3, \quad z=4 t $$

Step-by-Step Solution

Answer

The distance is \(\frac{3\sqrt{22}}{2}\).

Step 1: Identify given elements

We are given a point \((-1, 4, 3)\) and a line parameterized by \(x=10+4t\), \(y=-3\), and \(z=4t\). The direction vector of the line is \(\vec{d} = \langle 4, 0, 4 \rangle\), and a point on the line (when \(t=0\)) is \((10, -3, 0)\).

Step 2: Use the formula for distance

The distance \(d\) from a point \((x_1, y_1, z_1)\) to a line passing through a point \((x_0, y_0, z_0)\) with direction vector \((a, b, c)\) is given by:\[d = \frac{|\vec{AB} \times \vec{d}|}{|\vec{d}|}\]where \(\vec{AB}\) is the vector from the point on the line to the given point.

Step 3: Calculate \(\vec{AB}\)

To find \(\vec{AB}\), subtract the coordinates of the point on the line from the given point:\[\vec{AB} = \langle -1 - 10, 4 - (-3), 3 - 0 \rangle = \langle -11, 7, 3 \rangle\]

Step 4: Calculate \(\vec{AB} \times \vec{d}\)

Compute the cross product of \(\vec{AB}\) and \(\vec{d}\):\[\vec{AB} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -11 & 7 & 3 \ 4 & 0 & 4 \end{vmatrix}\]This simplifies to:\[\vec{AB} \times \vec{d} = \langle 28 - 0, 12 + 44, 0 + 44 \rangle = \langle 28, 56, 44 \rangle\]

Step 5: Calculate the magnitudes

First, find the magnitude of \(\vec{AB} \times \vec{d}\):\[|\vec{AB} \times \vec{d}| = \sqrt{28^2 + 56^2 + 44^2} = \sqrt{784 + 3136 + 1936} = \sqrt{5856}\]Next, find the magnitude of \(\vec{d}\):\[|\vec{d}| = \sqrt{4^2 + 0^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\]

Step 6: Compute the distance

Use the formula from Step 2 to find the distance:\[d = \frac{|\vec{AB} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{5856}}{4\sqrt{2}}\]Simplify this further:\[\sqrt{5856} = \sqrt{144 \times 11} = 12\sqrt{11}\]Hence the distance is:\[\frac{12\sqrt{11}}{4\sqrt{2}} = \frac{3\sqrt{11}}{\sqrt{2}} = \frac{3\sqrt{22}}{2}\]

Step 7: Summary of the solution

The distance from the point \((-1, 4, 3)\) to the line is \( \frac{3\sqrt{22}}{2} \).

Key Concepts

Direction Vector

The direction vector is a key component when dealing with lines in three-dimensional space. It indicates the direction in which a line extends. From the original equation of the line given, the parametric equations are:

• \(x = 10 + 4t\)
• \(y = -3\)
• \(z = 4t\)

If we look closely, the coefficients next to \(t\) are components of the direction vector, which in this case is \(\vec{d} = \langle4, 0, 4\rangle\). This vector shows that as \(t\) changes, the line moves 4 units in the x-direction, 0 units in the y-direction, and 4 units in the z-direction.
Thus, the direction vector not only tells you in which direction the line is heading but also provides insight into the slope of the line in space.

Cross Product

The cross product plays an important role when computing distances from a point to a line. It helps in finding a vector perpendicular to both vectors involved.
The formula for the cross product of two vectors \(\vec{u}\) and \(\vec{v}\), where \(\vec{u} = \langle u_1, u_2, u_3 \rangle\) and \(\vec{v} = \langle v_1, v_2, v_3 \rangle\), is given by the determinant:

\[\vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}\]Using this for vectors \(\vec{AB} = \langle -11, 7, 3 \rangle\) and \(\vec{d} = \langle 4, 0, 4 \rangle\) in our exercise, we compute:
• \(\vec{AB} \times \vec{d} = \langle 28, 56, 44 \rangle\)
The resulting vector \(\vec{AB} \times \vec{d}\) is orthogonal to the plane formed by \(\vec{AB}\) and \(\vec{d}\). This orthogonal property is crucial for calculating distances.

Magnitude Calculation

Calculating the magnitude of vectors is essential in various geometric computations, including finding the distance between a point and a line. The magnitude of a vector \(\vec{v} = \langle v_1, v_2, v_3 \rangle\) is given by:

\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]

In the original problem, we compute the magnitude of the cross product vector \(\vec{AB} \times \vec{d}\) as follows:

• \(|\vec{AB} \times \vec{d}| = \sqrt{28^2 + 56^2 + 44^2} = \sqrt{5856}\)

Additionally, the magnitude of the direction vector \(\vec{d}\) is:

• \(|\vec{d}| = \sqrt{4^2 + 0^2 + 4^2} = 4\sqrt{2}\)

These magnitudes are critical for computing the overall distance from the point to the line. They are cool tools in transforming geometric information into actionable insights.

Parametric Equations of a Line

Parametric equations are used to represent a line in three dimensions using an independent parameter, usually denoted as \(t\). They allow us to express each coordinate of a point on the line as a function of this parameter. For a line passing through a point \((x_0, y_0, z_0)\) with a direction vector \(\vec{d} = \langle a, b, c \rangle\), the parametric equations are:

• \(x = x_0 + at\)
• \(y = y_0 + bt\)
• \(z = z_0 + ct\)

In the exercise, the parametric equations \(x=10+4t\), \(y=-3\), and \(z=4t\) represent a line in 3D space. They help us identify a specific point on the line when \(t=0\), which is \((10, -3, 0)\).
Using these equations, we can explore various points on the line by varying \(t\), and understand its position and orientation in space. Parametric equations are essential for translating algebraic and geometric interpretations into comprehensible insights.